Taylor series.

Definition 10.5.1   Let $ f$ be a function having derivatives of any order at $ x_0$ . The power series

$\displaystyle P(x)$ $\displaystyle = \underset{n=0}{\overset{+ \infty}{\sum}} \frac {f^{(n)}(x_0)}{n!} (x-x_0)^n$    
$\displaystyle = f(x_0)+(x-x_0)f'(x_0)$ $\displaystyle +\frac {1}{2!}(x-x_0)^2f''(x_0)$    
$\displaystyle \quad$ $\displaystyle +\frac {1}{3!}(x-x_0)^3f'''(x_0)+\dots + \frac {1}{2!}(x-x_0)^nf^{(n)}(x_0) + \dots$    

is called the Taylor series of the function $ f$ at $ x_0$ .

If $ x_0=0$ , this series is called the Mac-Laurin series of $ f$ .

Example 10.5.2 (A couple of MacLaurin series)       

Theorem 10.5.3   Let $ f$ and $ g$ be two functions having Taylor series expansions $ \underset{n=0}{\overset{+ \infty}{\sum}} a_n x^n$ and $ \underset{n=0}{\overset{+ \infty}{\sum}} b_n x^n$ respectively, converging on the interval $ (-R,R)$ . Then $ f+g$ has as its Taylor series expansion the series $ \underset{n=0}{\overset{+ \infty}{\sum}} (a_n+b_n) x^n$ and it converges on $ (-R,R)$ .

Theorem 10.5.4   Let $ f$ and $ g$ be two functions having Taylor series expansions $ \underset{n=0}{\overset{+ \infty}{\sum}} a_n x^n$ and $ \underset{n=0}{\overset{+ \infty}{\sum}} b_n x^n$ respectively, converging on the interval $ (-R,R)$ . Then $ fg$ has as its Taylor series expansion the product of these series and it converges on $ (-R,R)$ .

Example 10.5.5   On the one side we have:

$\displaystyle \sin 2x = (2x) - \frac {(2x)^3}{3!} + \frac {(2x)^5}{5!} - \dots + (-1)^n \frac {(2x)^{2n+1}}{(2n+1)!} + \dots$    

On the other side, we have:

$\displaystyle \sin x$ $\displaystyle = x - \frac {x^3}{3!} + \frac {x^5}{5!} - \dots + (-1)^n \frac {x^{2n+1}}{(2n+1)!} + \dots$    
$\displaystyle \cos x$ $\displaystyle = 1 - \frac {x^2}{2!} + \frac {x^4}{4!} - \dots + (-1)^n \frac {x^{2n}}{(2n)!} + \dots$    
Thus, we have: $\displaystyle \quad$    
$\displaystyle 2 \sin x \cos x$ $\displaystyle = 2x - (\frac 12 + \frac 16) x^3 + ( \frac {1}{120} + \frac {1}{12} +\frac {1}{24})x^5 + \dots$    

Dear reader, please check that these two series are the same.

Theorem 10.5.6   Let $ f$ and $ g$ be two functions having Taylor series expansions in an open interval about 0. The Taylor series expansion of $ gof$ is obtained by substituting the Taylor series of $ f$ into the Taylor series of $ g$ .

Example 10.5.7   We have:

$\displaystyle \ln (1+x)$ $\displaystyle = x + \frac {x^2}{2} + \frac {x^3}{3} + \dots + \frac {x^n}{n} + \dots$    
$\displaystyle \sin x$ $\displaystyle = x - \frac {x^3}{3!} + \frac {x^5}{5!} - \dots + (-1)^n \frac {x^{2n+1}}{(2n+1)!} + \dots$    
Thus: $\displaystyle \quad$    
$\displaystyle \ln (1+ \sin x )$ $\displaystyle = x - \frac 12 x^2 + \frac 16 x^3 - \frac {1}{12} x^4 + \frac {1}{24} x^5 + \dots$    

Noah Dana-Picard 2007-12-28